OFFSET
1,3
COMMENTS
All terms are cubes. Proof: Let d_k be the k-th divisor of some odd m > 1 and t be the number of divisors of m. Then d_(t-1) is <= n/3 and so any sum of 3 divisors of at most d_(t-1) is at most n and so that sum is counted per A093035. A sum of 3 divisors of m where one of the divisors is d_t = m as more than m so not counted. This gives (t-1)^3 possible triples hence all terms are cubes.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David A. Corneth, May 19 2020
STATUS
approved